I need to prove $$\operatorname{det}\begin{pmatrix}A & B \\ C & D\\ \end{pmatrix}= \operatorname{det}(DA-CB),$$ where $A,B,C,D \in M_{n\times n}(R)$ with the property that $A$ and $B$ commute and moreover $\operatorname{det}(B) \neq 0,$ using the following hint: $$\begin{pmatrix}A & B \\ C & D\\ \end{pmatrix}\begin{pmatrix}B & 0 \\ -A & I_n\\ \end{pmatrix}.$$
Using the hint, I have: $$\begin{pmatrix}A & B \\ C & D\\ \end{pmatrix}\begin{pmatrix}B & 0 \\ -A & I_n\\ \end{pmatrix} = \begin{pmatrix}0 & B \\ CB-DA & D\\ \end{pmatrix}.$$Then:
$$\operatorname{det}\begin{pmatrix}A & B \\ C & D\\ \end{pmatrix}\operatorname{det}\begin{pmatrix}B & 0 \\ -A & I_n\\ \end{pmatrix} = \operatorname{det}\begin{pmatrix}0 & B \\ CB-DA & D\\ \end{pmatrix},$$ but $$\operatorname{det}\begin{pmatrix}B & 0 \\ -A & I_n\\ \end{pmatrix} = \operatorname{det}(B)$$ $$\operatorname{det}\begin{pmatrix}0 & B \\ CB-DA & D\\ \end{pmatrix} = (-1)^n\operatorname{det}(CB-DA)\operatorname{det}(B).$$ Then: $$\operatorname{det}\begin{pmatrix}A & B \\ C & D\\ \end{pmatrix}\operatorname{det}(B) = (-1)^n\operatorname{det}(CB-DA)\operatorname{det}(B)$$ which can be rewritten as follows:
$$[\operatorname{det}\begin{pmatrix}A & B \\ C & D\\ \end{pmatrix}+(-1)^{n+1}\operatorname{det}(CB-DA)]\operatorname{det}(B) = 0$$
But $\operatorname{det}(B) \neq 0$, hence:$$\operatorname{det}\begin{pmatrix}A & B \\ C & D\\ \end{pmatrix}+(-1)^{n+1}\operatorname{det}(CB-DA) = 0,$$ So
$$\operatorname{det}\begin{pmatrix}A & B \\ C & D\\ \end{pmatrix} = (-1)^n\operatorname{det}(CB-DA).$$
Im stuck on this part, can anyone help me on this step